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Cosine Fourier seires of cosh(t-1)

  1. Jul 12, 2006 #1
    Hi, can someone give me some help with the following?

    The cosine Fourier series of period 2 for the function f(t) that takes the form f(t) = cosh(t-1) in the range [itex]0 \le t \le 1[/itex] is

    [tex]
    \cosh \left( {t - 1} \right) = \sinh \left( 1 \right)\left[ {1 + 2\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {n\pi t} \right)}}{{n^2 \pi ^2 + 1}}} } \right]
    [/tex]

    Setting t = 0 yields

    [tex]\sum\limits_{n = 1}^\infty {\frac{1}{{n^2 \pi ^2 + 1}}} = \frac{1}{{e^2 - 1}}[/tex]...(1)

    I need to deduce the values of the sum [tex]\sum\limits_{}^{} {\left( {n^2 \pi ^2 + 1} \right)^{ - 1} } [/tex] over odd n and even n.

    I tried setting t = 1 so that I could evaluate the sum

    [tex]
    \sum\limits_{n = 1}^\infty {\frac{{\left( { - 1} \right)^n }}{{n^2 \pi ^2 + 1}}}
    [/tex]...(2)

    Once I did that, I added (2) to (1) and it looked like I would end up with just twice the sum of the even terms since the odd terms cancel. But that's not right since I don't get the correct answer when I divide the result by 2. Can someone suggest a way to do calculate the sum over odd n and over even n? Thanks.
     
  2. jcsd
  3. Jul 12, 2006 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Certainly sounds like it should work. What did you get (2) equal to and what did you get for the sum of even terms? Since you say "I didn't get the correct answer" I assume you have an answer key. What is the "correct answer"?
     
  4. Jul 12, 2006 #3
    I found that

    [tex]
    \sum\limits_{n = 1}^\infty {\frac{{\left( { - 1} \right)^n }}{{1 + \pi ^2 n^2 }}} = \frac{e}{{e^2 - 1}} - \frac{1}{2}
    [/tex]

    So that the required sum (over even n) as I thought it would be is:

    [tex]
    \frac{1}{2}\left[ {\sum\limits_{n = 1}^\infty {\frac{{\left( { - 1} \right)^n }}{{1 + \pi ^2 n^2 }}} + \sum\limits_{n = 1}^\infty {\frac{1}{{1 + \pi ^2 n^2 }}} } \right]
    [/tex]

    [tex]
    = \frac{1}{2}\left[ {\frac{e}{{e^2 - 1}} - \frac{1}{2} + \frac{1}{{e^2 - 1}}} \right]
    [/tex]

    [tex]
    = \frac{1}{2}\left[ {\frac{1}{{e - 1}} - \frac{1}{2}} \right]
    [/tex]

    [tex]
    = \frac{1}{2}\left( {\frac{{3 - e}}{{2\left( {e - 1} \right)}}} \right)
    [/tex]

    [tex]
    = \frac{{3 - e}}{{4\left( {e - 1} \right)}}
    [/tex]

    which is the given answer for the sum over even n. It looks like a made a transposition error (forgot about a "+1") after I susbtituted t = 1 in the Fourier series. Thanks for the help.
     
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